"Direct Proof" of the Steiner-Lehmus Theorem The Steiner-Lehmus Theorem is famous for its indirect proof. I wanted to come up with a 'direct' proof for it (of course, it can't be direct because some theorems used, will, of course, be indirect). I started with $\Delta ABC$,  with angle bisectors $BX$ and $CY$, and set them as equal. The first obvious step was the Stewart's Theorem: 
$a(p^2 + mn) = b^2m + c^2n$
Since an angle bisector divides the third side into the same ratio as the ratio of the other two sides, I set $m = kc$, $n = kb$ and $k = \frac{a}{b + c}$. I plugged them into the equation and got (after some manipulation):
$p^2 = bc\left(1 - \left(\frac{a}{b + c}\right)^2\right)$.
Now, $CY^2 = ab\left(1 - \left(\frac{c}{a + b}\right)^2\right)$ and $BX^2 = ac\left(1 - \left(\frac{b}{a + c}\right)^2\right)$
Since $CY$ = $BX$,the above equations are equal:
$b\left(1 - \left(\frac{c}{a + b}\right)^2\right) = c\left(1 - \left(\frac{b}{a + c}\right)^2\right)$
It is sufficient to prove that $\frac{c}{a+b} = \frac{b}{a+c}$.
Or, $\frac bc = \frac{a + c}{a + b}$
Which can be easily disproved by a counterexample. Did I make a mistake, or is my approach incorrect?
EDIT: Isn't this essentially disproving the theorem?
 A: Your method looks fine but I don't know what you mean by "easily disproved by a counterexample."
If $a,b,c>0$ then
$$
\frac{b}{c}=\frac{a+c}{a+b} \iff b^2+ab=c^2+ac \iff b=c
$$
so indeed it is sufficient to prove that. (There are algebraic counterexamples if you let $c=-a-b$, which seems to be why this type of proof is considered "indirect," but which is not normally allowed for the sides of a triangle.)
To be more explicit, you have the statements:
$$
\begin{array}{cl}
\mathbf A & \text{Given a triangle with sides }a,b,c\text{ and two equal angle bisectors.} \\
\mathbf B & 0<a<b+c, 0<b<a+c, 0<c<a+b \\
\mathbf C &b\left(1 - \left(\frac{c}{a + b}\right)^2\right) = c\left(1 - \left(\frac{b}{a + c}\right)^2\right) \\
\mathbf D & \frac{c}{a+b} = \frac{b}{a+c} \\
\mathbf E & b=c \\
\mathbf F & \text{The triangle is isosceles.}
\end{array}
$$
$\mathbf A \implies \mathbf B$ by the definition of "triangle," and $\mathbf E \implies \mathbf F$ by the definition of "isosceles."
The Steiner-Lehmus theorem is $\mathbf A \implies \mathbf F$.
What you have shown above is $\mathbf A \implies \mathbf C$.
The way I interpreted your comment "It is sufficient to prove that $\frac{c}{a+b} = \frac{b}{a+c}$" is that you intended to use that 
$$
\mathbf B \wedge \mathbf C \wedge \left(\frac{c}{a+b} = \frac{b}{a+c} = x\right)
\\
\implies (0<x<1) \wedge \left(b(1-x^2)=c(1-x^2)\right) \\
\implies b=c \implies \mathbf F
$$
that is, it is sufficient to somehow prove $\mathbf D$, after which
your proof will be
$$
\mathbf A \implies \mathbf B \wedge \mathbf C \\
\mathbf B \wedge \mathbf C \wedge \mathbf D \implies \mathbf E \implies \mathbf F
$$
It is true that this would be sufficient, but as I showed above $\mathbf B \implies (\mathbf D \iff \mathbf E)$ already, it doesn't simplify the proof because $\mathbf D$ only holds for isosceles triangles from the start.
Your counterexample $b=3,a=4,c=5$ shows only that $\mathbf D$ does not unconditionally follow from $\mathbf B$. But this doesn't disprove the theorem, because this triangle does not have two equal angle bisectors, so $\mathbf A$ is not satisfied.
